yourshail123 wrote:If x and y are integers, what is the remainder when (x^2 + y^2) divided by 5?
A) When (x - y) is divided by 5, the remainder is 1.
B) When (x + y) is divided by 5, the remainder is 2.
Statement 1:
x-y = 5a + 1 = 1, 6, 11, 36...
Since x and y can take on many different integer values, INSUFFICIENT.
Statement 2:
x+y = 5b + 2 = 2, 7, 12, 17...
Since x and y can take on many different integer values, INSUFFICIENT.
Statements combined:
x-y = 1, 6, 11, 21...
x+y = 2, 7, 12, 17...
(x-y) + (x+y) = 2x.
Since x is an integer, (x-y) + (x+y) must be even.
Thus, x-y and x+y are either BOTH EVEN or BOTH ODD.
Note the following:
(x-y)² + (x+y)² = (x² - 2xy + y²) + (x² + 2xy + y²) = 2(x² + y²).
(x-y)² = 1, 36, 121, 256...
(x+y)² = 4, 49, 144, 289...
If (x-y)² and (x+y)² are BOTH EVEN, their sum will have a units digit of 0:
36+4=40.
256+144=400.
A units digit of 0 implies a multiple of 10.
If (x-y)² and (x+y)² are BOTH ODD, their sum will have a units digit of 0:
1+49=50.
121+289=410.
A units digit of 0 implies a multiple of 10.
Thus, whether (x-y)² and (x+y)² are BOTH EVEN or BOTH ODD:
(x-y)² + (x+y)² = multiple of 10.
2(x² + y²) = multiple of 10
x² + y² = (multiple of 10)/2 = multiple of 5.
Thus, when x²+y² is divided by 5, the remainder will be 0.
SUFFICIENT.
The correct answer is
C.
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